Let $a_1 = 1$ and pick $a_n$ from $2, 3, ..., p$ such that $a_n\equiv \frac{n}{n-1} \mod{p}$ for $2 \le n \le p$. Then $a_1...a_n\equiv 1 \cdot \frac{2}{1} \cdot ... \cdot \frac{n}{n-1} \mod{p} \equiv n \mod{p}$ This satisfies the condition, so if we prove that $a_1, a_2, ..., a_p$ are distinct, then we are done. First, we prove that $a_n\equiv \frac{n}{n-1} \mod{p} \not\equiv 1 \mod{p}$ for $2 \le n \le p$. Assume there exists $n$ such that $a_n\equiv \frac{n}{n-1} \mod{p} \equiv 1 \mod{p}$. Then $n \equiv n-1 \mod{p} \Leftrightarrow 1 \equiv 0 \mod{p}$. But $p \ge 2$, so this is impossible. Second, we prove that $a_i \neq a_j$ for $2 \le i < j \le p$. Assume that $a_i = a_j$, so $a_i \equiv a_j \mod{p}$ $\frac{i}{i-1} \equiv \frac{j}{j-1} \mod{p}$ $ij - i \equiv ij - j \mod{p}$ $i \equiv j \mod{p}$, but $2 \le i < j \le p$, which is impossible again. Therefore, $a_1, a_2, ..., a_p$ are distinct, and we are done. $\blacksquare$

Another solution: pick up generator $g$ and consider sequence $b_{2i}=n-i, b_{2i-1}=i$ for $i\le \frac{p-1}{2}$. It is easy to see that $a_i=g^{b_i}$ and $a_p=p$ satisfy condition of the problem(so in fact we prove existence of $\phi (p-1)$ such sequences)

..here:)